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When a dart is thrown at a circular target, consider the location of the landing point relative to the bull鈥檚 eye. Let \({\rm{X}}\) be the angle in degrees measured from the horizontal, and assume that \({\rm{X}}\) is uniformly distributed on \(\left( {{\rm{0, 360}}} \right){\rm{.}}\)Define\({\rm{Y}}\)to be the transformed variable \({\rm{Y = h(X) = (2\pi /360)X - \pi ,}}\) so \({\rm{Y}}\) is the angle measured in radians and\({\rm{Y}}\)is between \({\rm{ - \pi and \pi }}\). Obtain \({\rm{E(Y)}}\)and\({{\rm{\sigma }}_{\rm{y}}}\)by first obtaining E(X) and \({{\rm{\sigma }}_{\rm{X}}}\), and then using the fact that \({\rm{h(X)}}\) is a linear function of \({\rm{X}}\).

Short Answer

Expert verified

\(\begin{array}{*{20}{c}}{{\rm{E(X) = 180;}}{{\rm{\sigma }}_{\rm{x}}}{\rm{ = 103}}{\rm{.9230}}}\\{{\rm{E(Y) = 0;}}{{\rm{\sigma }}_{\rm{y}}}{\rm{ = (0}}{\rm{.57735)\pi }}}\end{array}\)

Step by step solution

01

Definition of probability

The proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

02

 Calculating when a dart when it’s thrown at a circular target, consider the location of the landing point relative to the bull’s eye

Because \({\rm{X}}\)is uniformly distributed on \({\rm{(0,360)}}\), its pdf \({\rm{(x)}}\) can be written as follows for all \({\rm{x}}\) in this interval:

\({\rm{f(x) = }}\frac{{\rm{1}}}{{{\rm{360 - 0}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{360}}}}\)

Overall, \({\rm{f(x)}}\) can be written as: \({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{c}}{\frac{{\rm{1}}}{{{\rm{360}}}}{\rm{0\poundsx\pounds360}}}\\{{\rm{\;0 otherwise\;}}}\end{array}} \right.\)

\({\rm{X}}\)'s expected value can be expressed as: \(\begin{array}{*{20}{c}}{{\rm{E(X) = \`o }}_{{\rm{ - \currency}}}^{\rm{\currency}}{\rm{nx \times f(x) \times dx}}}\\{{\rm{ = \`o }}_{\rm{0}}^{{\rm{360}}}{\rm{nx \times }}\frac{{\rm{1}}}{{{\rm{360}}}}{\rm{ \times dx}}}\\{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{360}}}}{\rm{\`o }}_{\rm{0}}^{{\rm{360}}}{\rm{nx \times dx}}}\\{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{360}}}}\left( {\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}} \right)_{\rm{0}}^{{\rm{360}}}}\\{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{720}}}}\left( {{\rm{36}}{{\rm{0}}^{\rm{2}}}{\rm{ - }}{{\rm{0}}^{\rm{2}}}} \right)}\\{{\rm{E(X) = 180}}}\end{array}\)

Definition: A continuous \({\rm{rv's}}\) expected or mean value. \({\rm{f(x)}}\) is \({\rm{X}}\)with pdf

\({\rm{E}}\left( {{{\rm{X}}^{}}} \right)\)\({\rm{ = }}\mathop {\rm{\`o }}\nolimits_{{\rm{ - \currency}}}^{\rm{\currency}} {{\rm{x}}^{\rm{2}}}{\rm{ \times f(x) \times dx}}\)

Now consider the following argument:

The standard deviation \({{\rm{\sigma }}_{\rm{x}}}\) and variance \({\rm{V(X)}}\)of a \({\rm{rv's}}\) \({\rm{X}}\)with a specified pdf can be stated as: \(\begin{array}{*{20}{c}}{{\rm{V(X) = E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ - (E(X)}}{{\rm{)}}^{\rm{2}}}}\\{{{\rm{\sigma }}_{\rm{x}}}{\rm{ = }}\sqrt {{\rm{V(X)}}} }\end{array}\)

We begin by calculating \({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)\).

\(\begin{array}{*{20}{c}}{{\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)}&{{\rm{ = }}\mathop {\rm{\`o }}\nolimits_{{\rm{ - \currency}}}^{\rm{\currency}} {{\rm{x}}^{\rm{2}}}{\rm{ \times f(x) \times dx}}}\\{}&{{\rm{ = }}\mathop {\rm{\`o }}\nolimits_{\rm{0}}^{{\rm{360}}} {{\rm{x}}^{\rm{2}}}{\rm{ \times }}\frac{{\rm{1}}}{{{\rm{360}}}}{\rm{ \times dx}}}\\{}&{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{360}}}}\mathop {\rm{\`o }}\nolimits_{\rm{0}}^{{\rm{360}}} {{\rm{x}}^{\rm{2}}}{\rm{ \times dx}}}\\{}&{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{360}}}}\left( {\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right)_{\rm{0}}^{{\rm{360}}}}\\{}&{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{1080}}}}\left( {{\rm{36}}{{\rm{0}}^{\rm{3}}}{\rm{ - }}{{\rm{0}}^{\rm{2}}}} \right)}\\{{\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)}&{{\rm{ = 43200}}}\end{array}\)

03

Determining the probability using proprsition

Using the above proposition, we can now write:

\(\begin{array}{*{20}{c}}{{\sigma _x}}&{ = \sqrt {V(X)} }\\{}&{ = \sqrt {E\left( {{X^2}} \right) - {{(E(X))}^2}} }\\{}&{\sqrt {43200 - {{(180)}^2}} }\end{array}\)

\(\sqrt {43200 - 32400} \)

\(\sqrt {10800} \)

\({\sigma _x} = 103.9230\)

We can apply the following proposition because \({\rm{Y( = h(X)}}\) is a linear function of \({\rm{X}}\):

The expected value and standard deviation of \({\rm{h(X)}}\) meet the following conditions for \({\rm{h(X) = aX + b}}\).

\(\begin{array}{*{20}{c}}{{\rm{E(h(X)) = aE(X) + b}}}\\{{{\rm{\sigma }}_{{\rm{h(x)}}}}{\rm{ = a}}{{\rm{\sigma }}_{\rm{x}}}}\end{array}\)

We are informed that:

\({\rm{Y = }}\frac{{{\rm{2\pi }}}}{{{\rm{360}}}}{\rm{ \times X - \pi }}\)

As a result, the predicted value of \({\rm{Y}}\) is:

\(\begin{array}{*{20}{c}}{{\rm{E(Y)}}}&{{\rm{ = }}\frac{{{\rm{2\pi }}}}{{{\rm{360}}}}{\rm{ \times E(X) - \pi }}}\\{}&{{\rm{ = }}\frac{{{\rm{2\pi }}}}{{{\rm{360}}}}{\rm{ \times 180 - \pi }}}\\{}&{{\rm{ = }}\frac{{{\rm{360}}}}{{{\rm{360}}}}{\rm{ \times \pi - \pi }}}\\{{\rm{E(Y)}}}&{{\rm{ = 0}}}\end{array}\)

The standard deviation of \({\rm{Y}}\) is calculated as follows:

\(\begin{array}{*{20}{c}}{{{\rm{\sigma }}_{\rm{y}}}{\rm{ = }}\frac{{{\rm{2\pi }}}}{{{\rm{360}}}}{\rm{ \times }}{{\rm{\sigma }}_{\rm{x}}}}\\{{\rm{ = }}\frac{{{\rm{2\pi }}}}{{{\rm{360}}}}{\rm{ \times (103}}{\rm{.923)}}}\\{{{\rm{\sigma }}_{\rm{y}}}{\rm{ = (0}}{\rm{.57735)\pi }}}\end{array}\)

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Most popular questions from this chapter

In response to concerns about nutritional contents of fast foods, McDonald鈥檚 has announced that it will use a new cooking oil for its French fries that will decrease substantially trans fatty acid levels and increase the amount of more beneficial polyunsaturated fat. The company claims that \({\bf{97}}\)out of \({\bf{100}}\) people cannot detect a difference in taste between the new and old oils. Assuming that this figure is correct (as a long-run proportion), what is the approximate probability that in a random sample of \({\bf{1000}}\) individuals who have purchased fries at McDonald鈥檚, a. At least \({\bf{40}}\) can taste the difference between the two oils? b. At most \({\bf{5}}\% \)can taste the difference between the two oils?

Let \({\rm{X}}\) be a continuous \({\rm{rv}}\) with cdf

\({\rm{F(x) = }}\left\{ {\begin{array}{*{20}{c}}{\rm{0}}&{{\rm{x}} \le {\rm{0}}}\\{\frac{{\rm{x}}}{{\rm{4}}}\left( {{\rm{1 + ln}}\left( {\frac{{\rm{4}}}{{\rm{x}}}} \right)} \right)}&{{\rm{0 < x}} \le {\rm{4}}}\\{\rm{1}}&{{\rm{x > 4}}}\end{array}} \right.\)

(This type of cdf is suggested in the article 鈥淰ariability in Measured Bedload Transport Rates鈥 (Water Resources Bull., \({\rm{1985:39 - 48}}\)) as a model for a certain hydrologic variable.) What is a. \({\rm{P(X}} \le {\rm{1)}}\)? b. \({\rm{P(1}} \le {\rm{X}} \le {\rm{3)}}\)? c. The pdf of \({\rm{X}}\)?

a. Show that if X has a normal distribution with parameters \({\rm{\mu }}\) and\({\rm{\sigma }}\), then \({\rm{Y = aX + b}}\) (a linear function of X ) also has a normal distribution. What are the parameters of the distribution of Y (i.e., E(Y) and V(Y)) ? (Hint: Write the cdf of\({\rm{Y,P(Y}} \le {\rm{y)}}\), as an integral involving the pdf of X, and then differentiate with respect to y to get the pdf of Y.)

b. If, when measured in\(^{\rm{^\circ }}{\rm{C}}\), temperature is normally distributed with mean 115 and standard deviation 2 , what can be said about the distribution of temperature measured in\(^{\rm{^\circ }}{\rm{F}}\)?

The article 鈥淢odelling Sediment and Water Column Interactions for Hydrophobic Pollutants鈥 (Water Research, \({\rm{1984: 1169 - 1174}}\)) suggests the uniform distribution on the interval \({\rm{(7}}{\rm{.5,20)}}\)as a model for depth (cm) of the bioturbation layer in sediment in a certain region. a. What are the mean and variance of depth? b. What is the cdf of depth? c. What is the probability that observed depth is at most \({\rm{10}}\)? Between \({\rm{10}}\) and \({\rm{15}}\)? d. What is the probability that the observed depth is within \({\rm{1}}\) standard deviation of the mean value? Within \({\rm{2 }}\) standard deviations?

The accompanying normal probability plot was constructed from a sample of \({\rm{30}}\) readings on tension for mesh screens behind the surface of video display tubes used in computer monitors. Does it appear plausible that the tension distribution is normal?

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